"""
难度：中等
给定一个候选人编号的集合 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的每个数字在每个组合中只能使用 一次 。
注意：解集不能包含重复的组合。 
示例 1:
输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
示例 2:
输入: candidates = [2,5,2,1,2], target = 5,
输出:
[
[1,2,2],
[5]
]
"""

class Solution:
    res = []
    path = []
    def backtrack(self, candidates: List[int], target: int, sum: int, start_index: int):
        if sum > target:
            return
        if sum == target:
            self.res.append(self.path[:])
            return

        for i in range(start_index, len(candidates)):
            if sum + candidates[i] > target:
                break
            if i > start_index and candidates[i] == candidates[i - 1]:
                continue
            sum += candidates[i]
            self.path.append(candidates[i])
            self.backtrack(candidates, target, sum, i + 1)
            sum -= candidates[i]
            self.path.pop()

    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        self.res.clear()
        self.path.clear()
        candidates.sort()
        self.backtrack(candidates, target, 0, 0)
        return self.res